The price for admission to an amusement park is different for the children and adults. They offer two specials as a savings to families. For 3 children and 2 adults, the price is $21. For 2 children and 1 adult the price is $12.
a. write a system of equations using c as the variable for the price for children and a as the variable for the price for adults. solve this system of equations.
b. what is the price for children's admission?for adult admission?
2007-01-27 14:39:32 · 12 answers · asked by jenjen 3 in Science & Mathematics ➔ Mathematics
a.
3c+2a=21----->i
2c+1a=12----->ii
b.
i-2*ii--> c=3
.......a=6
2007-01-27 14:46:02 · answer #1 · answered by Tharu 3 · 1⤊ 2⤋
3c+2a=21
2c+(1)a=12
You can use the "substitution method" to solve the system of equations by getting either a or c by themself to equal the rest of one of the equations. For example, use the second equation by subtracting 2c from both sides to get (1)a=12-2c. Then use what you know a is equal to in place of a in the first equation. This would make it 3c+2(12-2c)=21. Distribute to get 3c+24-4c=21, and simplify to get 24-c=21. Then, solve the equation by adding c to both sides to get 24=21+c, and subtracting 21 from both sides to get c=3.
Now, you know that c=3, and you had originally made the variable c represent children, so you know that the cost of children's admission is 3 dollars. To get the cost of adult admission, or a, plug your c=3 into one of the equations and solve it. So let's say we use 3c+2a=21. We would put 3 into the spot of c to get 3(3)+2a=21. Distribute to get 9+2a=21, and subtract 9 from each side to get 2a=12. Then, divide each side by 2 to get a=6. You now know that the cost of adult admission is 16 dollars because you had originally made the variable a represent the cost of adult admission.
I hope this helps! It sounds a lot more complicated than it really is. Just ask your math teacher if you don't understand how the whole process works. Good luck!
2007-01-27 22:59:57 · answer #2 · answered by hippomaiden 3 · 0⤊ 1⤋
wow this might just be 6th grade math haha.
a. 3c + 2a = 21
2c + a = 12
b. use substitution... rewrite the second equation so it says a = 12 - 2c
then substitute (12 - 2c) for a in the first equation so it says 3c + 2 (12 - 2c) = 21
then solve for c, which ends up equaling 3
plug in 3 for c into the second equation, and this means that a equals 6.
adults cost 6 dollars
children cost 3 dollars
2007-01-27 22:57:48 · answer #3 · answered by i <3 andy roddick 3 · 0⤊ 1⤋
3c + 2a = $21...........(1)
2c + a = $12 ...........(2)
From (2) a = $12 - 2c
Sub this in eq (1) 3c + 2($12 - 2c) = $21
-c = $21 - $24
-c = -$3
c=$3
Sub this value in (2)
2($3) + a = $12
a = $12 - $6
a = $6
2007-01-27 23:14:08 · answer #4 · answered by flying_phoenix 2 · 0⤊ 1⤋
kids =c
adults=a
3c+2a=21
2c+1a=12
(3 2) x= (21 2) y=(3 21)
(2 1) (12 1) (2 12) the 2( are connected
3*1+2*2=7
21*1+12*2=45
3*12+21*2=78
45/7=5
kid's =5
78/7=11.143
adults=11.14
2007-01-27 22:59:59 · answer #5 · answered by Jeff O 1 · 0⤊ 2⤋
Adults are $6 Children are $3.
3C+2A=21 2C+1A=12
What to do next . . . I dont know.
I do most math in my head, so I can't explain.
2007-01-27 22:49:58 · answer #6 · answered by shae b 1 · 0⤊ 1⤋
I calculated the answer to "b" as the price for children is $3.00 each & adults is $6.00 each.
I'm not quite sure what you are looking for for question "a" but maybe this will help you get there...
2007-01-27 22:59:31 · answer #7 · answered by Maria C 2 · 0⤊ 1⤋
3c+2a=21
2c+1a=12
so then a=12-2c (i mixed up the second equation)
put that in the first equation
and it's 3c+ 2(12-2c)=21
3c+24-4c=21
-c=-3
c=3
children's tickets are three dollars
adult's tickets are six dollars.
hope i could help =]
2007-01-27 22:50:46 · answer #8 · answered by Kelly . 2 · 0⤊ 1⤋
a.
3c + 2a = 21
2c + 1a = 12
b.
4c + 2a = 24
3c + 2a = 21
c = $3
a = 12 - 2*3
a = $6
2007-01-27 23:06:10 · answer #9 · answered by Helmut 7 · 0⤊ 1⤋
man do ur work by urself
2007-01-27 22:50:11 · answer #10 · answered by iceboy_cripz 1 · 0⤊ 1⤋